2020 State Competition Solutions - Mathcounts
2020 State Competition SolutionsAre you wondering how we could have possibly thought that a Mathlete would beable to answer a particular Sprint Round problem without a calculator?Are you wondering how we could have possibly thought that a Mathlete would beable to answer a particular Target Round problem in less 3 minutes?Are you wondering how we could have possibly thought that a particular TeamRound problem would be solved by a team of only four Mathletes?The following pages provide solutions to the Sprint, Target and Team Rounds ofthe 2020 MATHCOUNTS State Competition. These solutions provide creative andconcise ways of solving the problems from the competition.There are certainly numerous other solutions that also lead to the correct answer,some even more creative and more concise!We encourage you to find a variety of approaches to solving these fun andchallenging MATHCOUNTS problems.Special thanks to solutions authorHoward Ludwigfor graciously and voluntarily sharing his solutionswith the MATHCOUNTS community.
2020 State Competition Sprint Round Solutions1. The angle referred to initially as B is more specifically ABC after the bisection. Due to the bisection,1ππ DBC ππ ABC 100/2 50 degrees. Because the sum of the measures of the three angles of a2triangle is always 180 , we have:180 ππ DBC ππ BCD ππ BDC 50 20 ππ BDC 70 ππ BDC, so:ππ BDC 180 70 110 degrees.2. We can divide both the numerator and the denominator by the numerator 1.2 102 , resulting in anumerator of 1 (dividing any nonzero number by itself yields 1), and a denominator of4 105 2 4 103 4000. Thus, we end up with the simplification1.2 1024.8 105 ππππππππππ.4.8 1051.2 102 3. I will use ππ instead of the airplane symbol and ππ instead of the ball symbol, making our two equations:ππ ππ 14;ππ ππ 4. Subtracting this lower equation from the above equation yields:2ππ 10, so ππ ππ.4. Starting at A we go 4 units right to get to B, then 2 units up to get to C, then 2 units left to get to D, then 2units up to get to E, then 2 units left to get to F, and finally 4 units down to get back to the starting point A.The total number of units traveled is 4 2 2 2 2 4 16.5. We want to partition the 40 ft 72 ft play area into squares of size ππ ft ππ ft. To avoid gaps and overlapwithout cutting tiles, both 40 and 72 must be divisible by ππ; since we want the largest possible tiles, weneed ππ to be the greatest common divisor of 40 and 72, which is 8. We will need 40/8 5 rows each with72/8 9 tiles for a total tile count of 5 9 45 square tiles.6. Let π π and ππ be the distance that Rose and Robyn, respectively, ran. We are given:π π ππ 29 mi and ππ 2π π β 4 mi, with the latter equation being equivalent to2π π ππ 4 mi. Since we want π π and not ππ, letβs add the two equations to get:3π π 33 mi, so that π π 11 miles.7. The problem does not state whether the two folds are parallel or perpendicular to each other, nor ifparallel whether the fold is along the short or the long edge of the original paper, so we consider alloptions. If the two folds are perpendicular, then each dimension of the final paper is 1/2 of the original,making the original sheet 8 in 10 in, with a perimeter double the sum of the two side lengths for 36 in.If the two folds are parallel, the effect of the two folds is to make one side the paper 1/4 as along as theoriginal while keeping the other side unchanged, making the original sheet either 4 in 20 in or16 in 5 in, depending on which original side the folds were made, resulting in a perimeter of 48 in or42 in, respectively. The greatest of these three possible values is 48 inches.8. In order for ππ! to have 7 factors of 2, we must have 8 ππ 9; in order to have 2 factors of 3, we must have6 ππ 8. These two pairs of constraints together tell us that ππ 8, which is consistent with havingexactly 1 factor of each of 5 and 7.9. 200(200 π₯π₯) 2072 49 2072 72 (207 7)(207 7) (214)200 200(200 14) so π₯π₯ 14.10. Letβs put the 0 point of our number line at A and put B on the positive side of A, thus at 24. C is at themidpoint of AB, thus 12. E is at the midpoint of CB, thus12 242 18. D is at the midpoint of AE, thus at
0 182 9. F is at the midpoint of AC, thusbetween F and G is 21 6 ππππ units.0 122 6. G is at the midpoint of EB, thus18 242 21. The distance11. Let ππ be the duration of a long movie and π π be the duration of a short movie. The total time duration of themovie watching was 5 hours 15 minutes (5 60 15) minutes 315 minutes. Other given32information includes: ππ 2π π 315 minutes and ππ 1.50π π π π , so π π ππ.Thus, 315 min ππ 4ππ3 7ππ3and ππ 3(315)72 3(45) ππππππ minutes.3322312. ππ (1 0.10)ππ 1.10ππ; ππ (1 0.25)ππ 0.75ππ; ππ (1 0.50)ππ 1.50ππ ππ so ππ ππ. We needππ ππas a percentage. Thus, ππ 1.1ππ 1.1(0.75ππ)ππππ 0.55ππ0.45ππ 0.45 0.45 100 % ππππ %.ππππto determineThenππ ππππ 23 1.1 0.75 ππ 1.1(0.5ππ) 0.55ππ.13. (π΄π΄ πΊπΊ) (π΄π΄ πΊπΊ) 2πΊπΊ 2 πΊπΊ .π΄π΄ (π΄π΄ πΉπΉ πΊπΊ) (πΉπΉ πΊπΊ) 11 10 1.π΅π΅ πΆπΆ (π΄π΄ π΅π΅ πΆπΆ) π΄π΄ 5 1 4.π·π· (π΅π΅ πΆπΆ π·π·) (π΅π΅ πΆπΆ) 7 4 3.πΈπΈ πΉπΉ (π·π· πΈπΈ πΉπΉ) π·π· 11 3 8.πΊπΊ (πΈπΈ πΉπΉ πΊπΊ) (πΈπΈ πΉπΉ) 13 8 5.Therefore, the desired value is 2 5 ππππ.14. To get from square 4 to square 32 using only downward motions, one must pass through exactly one ofsquares 18, 19 and 20. For each of those 3 options, we can determine how many ways lead from square 4to the square in question, and then how many ways lead from the square in question to square 32; sincethe ways from square 4 to the intermediate square are independent of the ways from the intermediatesquare to square 32, the number of ways from square 4 through the intermediate square underassessment to square 32 is the product of those two values. The total count of ways is the sum of thoseproducts over the 3 intermediate squares being examined. For square 18 there is 1 way from square 4(4.8.11.15.18) and 1 way to square 32 (18.23.27.32). For square 19 there are 3 ways from square 4(4.8.11.15.19; 4.8.11.16.19; 4.8.12.16.19) and 3 ways to square 32 (19.23.27.32; 19.24.27.32;19.24.28.32). For square 20 there are 2 ways from square 4 (4.8.11.16.20; 4.8.12.16.20) and 2 waysto square 32 (20.24.27.32; 20.24.28.32). Therefore, the total number of qualifying ways from square 4to square 32 is 1 1 3 3 2 2 ππππ ways.15. We have an uncut length, an uncut width, a shortened length, a shortened width, and a hypotenuse of theright triangular cutoff representing the values 10, 2, 11, 10 and 5 [not in the same order]. Because allvalues are integers, the cutoff piece sides need to be a Pythagorean triple. The only options for thehypotenuse are 5 (3:4:5) and 10 (6:8:10). A hypotenuse of 5 does not work because there are no two sidesdiffering by 3 and no two sides differing by 4; 10 does work because 11 5 6 and 10 2 8. Therefore,we have an 11 10 rectangle with a right triangular piece with legs of 6 and 8 removed, so the area of the6 8 110 24 ππππ units2.pentagon is 11 10 2
16. The arithmetic mean is the sum divided by the count, so π΄π΄ 200.9The median of 9 values in increasingorder is the fifth value. To allow the fifth value to be larger with a fixed sum of 200, letβs make the 4 smallestvalues as small as allowed [nonnegative], namely 0, and the 4 largest values as small as possible, so theyshould be equal to the fifth value. Thus, we have 4 values of 0 and 5 values that are equal and whose sumis 200. Thus, each of those values must be2005 40, so ππ 40 and ππ π΄π΄ 40 2009 360 2009 ππππππ.ππ17. The base of the pyramid will be the 3 congruent original outer edges of the square, thus forming anequilateral triangle with sides whose length is 2 times the length of each dotted segment, for 4 2 in. Thearea enclosed by an equilateral triangle with side π π is 3 2π π 4 2 3 4 2 4 ππ ππ in2 .18. (π₯π₯ 3 π₯π₯ 2 3π₯π₯ ππ)(2π₯π₯ 4 πππ₯π₯ 3 π₯π₯ 2 π₯π₯ 1) 2π₯π₯ 7 8π₯π₯ 6 π₯π₯ 5 4π₯π₯ 4 39π₯π₯ 3 11π₯π₯ 2 10π₯π₯ 7.Substitute two values for π₯π₯ to be able to solve for ππ and ππβkeep the values simple. Two values that arecommonly good to use for π₯π₯ are 0 and 1 in such problems:π₯π₯ 0: (ππ)(1) 7, so ππ 7.π₯π₯ 1: (6)(3 ππ) 54, so ππ 6.Therefore, ππ ππ ππππ.ππ19. ππ0 10 101 ; ππ1 102 ; ππ2 104 ; ππ3 108 ; . In general, ππππ 102 .123ππ123ππTherefore, the product of ππ1 through ππππ is 102 102 102 102 102 2 2 2 .The exponent in the last expression is a geometric series with first term 2, last term 2ππ , and common ratio2, so the sum is2 2ππ 11 2 2ππ 1 2. This makes the product to be 102ππ 1 2. Now, 10ππ has ππ 1 digits. Sincewe want at least 100 digits, the exponent needs to be 1 less. Therefore, 2ππ 1 2 99, so 2ππ 1 101. Thepowers of 2 in that region are 26 64 and 27 128, so ππ 1 7. The least value that works for ππ is 6.20. 1024 210 (25 )2 322 (22 )5 45 . Thus, there are 32 perfect squares and 4 perfect fifth powers for1 to 1024, inclusive. Of these, 2 values overlap in the two sets: 110 1 and 210 1024. (A number is botha perfect square and a perfect fifth power if and only if it is a perfect tenth power.) Thus, we start off with1024 integers, subtract 32 as being perfect squares and subtract 4 more as being perfect fifth powers,leaving 988 to consider. However, in doing so, we have deleted 1 and 1024 twice, so we need to add 2, tomake sure any deleted value is deleted exactly once. Thus, there are 990 such positive integers.21. We would like to be able to use all 10 digits in our number, and the leftmost digit should be a 9, if possible,to achieve the largest qualifying 10-digit number. Letβs start with 9. The next digit needs to divide 9 or bedivisible by 9βonly 3, 1 and 0 work, so try the largest, 3. Then the next digit needs to divide 3 or bedivisible by (and not already be used)βonly 6, 1 and 0 work, so try the largest, 6. Then the next digit mustbe 2, 1 or 0, so try the largest, 2. The next digit needs to be 8, 4, 1 or 0, so try the largest, 8. The next digitneeds to be 4, 1 or, 0, so try the largest, 4. The next digit needs to be 1 or 0, so try the largest, 1. Since 1divides anything, it will work we whatever we have left: 7, 5 or 0, so try the largest, 7. Only 0 and 5 remain,and, of these two, 7 divides only 0, so that must come next. For the last digit, 5 will work, because 5 divides0. We have successfully completed the largest possible qualifying number: 9,362,841,705.22. [1]: π₯π₯ π¦π¦ ππ[2]: 3π₯π₯ 2π¦π¦ ππ 15[3]: 4π₯π₯ 5π¦π¦ ππ 19Notice how adding equations [2] and [3] yields:[4]: 7π₯π₯ 7π¦π¦ 2ππ 34, in which the coefficients of π₯π₯ and π¦π¦ are equal, just as in [1].Letβs subtract 7 times equation [1] from equation [4] to yield:[5]: 0 5ππ 34, so ππ ππππ.ππ
23. There are 5 digits to permute, so 5! 120 total permutations, with 13,579 being the 1st and 97,531 beingthe 120th in increasing order by size of number. The leftmost digit is the one used for comparison first todecide which of two numbers is greater. Once the leftmost digit has been chosen for a given permutation,there are 4 digits remaining to permute, for a total of 4! 24 permutations. Thus, there are 24permutations for each possible leftmost digit. Thus, permutations 1 through 24 start with 1, 25 through48 start with 3, 49 through 72 start with 5, 73 through 96 start with 7, and 97 through 120 start with 9.Since we want the 100th permutation, that means we start with 9. For each choice of digit we might choosesecond, we will have 3! 6 permutations of the remaining 3 digits. Thus, permutations 97 through 102start with 91βthis includes our desired 100th permutation. We keep iterating this process: permutations97 and 98 start with 913, while 99 and 100 (which is what we want) start with 915. Then we want thesecond of 37 and 73, so we end up with 91,573.24. Note that each digit A through G occurs once in each of the 1s, the 10s, the 100s and the 1000s. That meansfor each place value we will end up with A B C D E F G, even though not necessarily occurringin that order, addition is commutative and associative. We will get 1000 of these for the leftmost columnof digits, 100 for the second column, 10 for the third, and 1 for the fourth. Thus the total sum will be:(1000 100 10 1)(A B C D E F G) 1111(A B C D E F G).Now, 1111 11 101, with 11 and 101 both prime, so 101 is the larger; the sum of the 7 digit valuescannot exceed 7 9 63, so this part contributes no prime factor greater than 101. Thus, the greatestprime factor of the sum of the four numbers must be 101, regardless the value of each letter representingsome digit 0 through 9.25. The figure shows the 12th triangular number of lattice points, thus12(12 1)2 78. However, we must read the question very carefully. Only βthe latticepoints inside the triangleβ [my italic emphasis added] are to be considered,that is, the blue and green points in the figure. This eliminates the 33 pointson the triangle (technically, the definition of a triangle includes only the3 sides and not the interior, though we often speak loosely of βthe area of atriangleβ, when we really mean the area enclosed by a triangle). Therefore,45 points are under consideration for selection. The triangle inequalityrequires that the longest side of a triangle must have length less than halfthe sum of the lengths of all 3 sides. In an equilateral triangle the sum of the distances from any one pointinterior to the triangle to the 3 sides of the triangle equals the altitude of the triangle [Vivianiβs theorem].Therefore, a point is desired if and only if for each vertex of the triangle, the point is farther from the vertexthan from the opposite side. This region of points is interior to the triangle formed with vertices at themidpoints of the sides of the original triangle, and the qualifying points are colored green in the figure.The probability is the number of green points divided by the number of points that are blue or green.15ππThere are 15 points interior to the new triangle, making the probability 45 ππ . [NOTE: The 4 trianglesformed by the 3 auxiliary lines are congruent, so if we were working in continuous space, the probability14of any (π₯π₯, π¦π¦) point being within the middle smaller equilateral triangle would be only ; however, wecannot assume with a discrete lattice that the lattice points distribute evenly across the differentsubregions, even though they might enclose the same physical area.]
26. The circle being tangent to the π₯π₯-axis at π₯π₯ 3 means that a diameter of the circle is perpendicular to theπ₯π₯-axis at π₯π₯ 3, meaning that the π₯π₯-component of the circle center is 3, and, when combined with theknowledge that the circle is at or above the π₯π₯-axis, the π¦π¦-component is equal to the radius π π of the circle.Thus, the equation of the circle is of the form (π₯π₯ 3)2 (π¦π¦ π π )2 π π 2 , so (π₯π₯ 3)2 π¦π¦ 2 2π π π π 0.Substitute a helpful (π₯π₯, π¦π¦) coordinate of a known point on the circle to determine the radius: we know(3; 0) is on the circle, but that is not helpful as we need π¦π¦ 0 to determine π π , so use (6; 6):4515150 9 36 12π π , so π π , making the circle have the equation (π₯π₯ 3)2 π¦π¦ 2 π¦π¦ 0. Another1242point on the circle is (ππ, ππ), so letβs substitute ππ for π₯π₯ and π¦π¦ in the equation and solve for ππ:1515270 (ππ 3)2 ππ2 ππ ππ2 6ππ 9 ππ2 ππ 2ππ2 ππ 9. Before applying the quadratic222formula, multiply the equation by 2 to simplify the fractions to be handled: 4ππ2 27ππ 18 0, which hassolutions ππ 27 729 2888 27 21.8ππππThe yields 6, which we were already given; the yields .234527. The probability of an integer being not divisible by 3 is , of being not divisible by 5 is , and of not being6divisible by 7 is 7 . Over a count of consecutive integers that is a multiple of the least common multiple of3, 5 and 7, these three probabilities are independent, so the probability that an integer in such a band is24616not divisible by any of 3, 5 and 7 is . The least common multiple of 3, 5, and 7 is the product35735of the three values, so 105. Thus, out of any 105 consecutive integers, exactly1635 105 48 will fail to bedivisible by each of 3, 5, and 7. That means that in the range 1 through 19 105 1995, there are exactly19 48 912 such integers. Itemize the remaining values and cull out those divisible by 3, 5 or 7:1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 20122013 2014 2015 2016 2017 2018 2019.There remain 11 values not shaded out, so the answer is 912 11 ππππππ positive integers.28. The only calculator you are allowed to use is the one inside your skull augmented with pencil and scratchpaper. The product has many digits (16) but its value is not that hard to determine given the values of thefactors. The hard part would be to extract the square root of a 16-digit integer, even though it is a perfectsquare. This is MATHCOUNTS, so if the answer does not say what form to use, it has to be an integer, so ifwe can make good, easy approximations and get close to an expected form with expected properties and1value, then the nearest integer of that form is very likely the answer. As 0.111 , with the 1s repeating918forever, so 11,111,111 is close to [slightly less than] 10 , while 100 000 011 is close to [slightly greater911816than] 10 , so the product is close to 9 10 . The square root must be close to 3 108 33,333,333. Thelast 2 digits of the product come from 11 11 121, thus 21 [and an electronic calculator would be verylikely be unable to tell you that]; adding the 4 yields 25, which is compatible with the number whosesquare root to be extracted having a units digit of 5. An excellent guess is 33,333,335. If you have time tospare, you can enhance your confidence in this guess by noticing your estimate for the left factor is too1919high by , which out of 108 is a relative error of only 1 part in 108 . Your estimate of 108 for 100,000,011is 11 too low for a relative error of 11 parts in 108 . The relative error of a product is close to the sum ofthe relative errors of the factors for small relative errors, so you are close to 10 parts in 108 . Adding 4 toa 16-digit number has a negligible effect on relative error. Finally, taking a square root cuts the relativeerror in half to 5 parts in 108 . Thus, you expect your answer to be very close to 5 parts in 108 too low,which would be510813531323 108 , so we need to raise our estimated answer of 33,333,333 by about 1 ,which results in our predicted 33,333,335.
29. The triangle is isosceles with base of length 6 and two congruent sides of length 5, so it can be partitionedinto to congruent right triangles with the two congruent sides (length 5) being the hypotenuse of the tworight triangles and the two halves of the base (length 3) each being one leg of the two right triangles. Thus,the two right triangles are 3:4:5 right triangles, so the height of the dartboard is the length of the secondleg, 4. By symmetry we can tell the minimum sum of the squares of the three distances will occur at thecentroid of the dartboard, so that the origin of the coordinate system should be put at the centroid of thedartboard. (Note: It is not critical to know to put the origin of the coordinate system at the centroid, but itavoids some completions of squares to solve that slow you down a bit.] To make solving the problem fasterwe will place the π₯π₯-axis parallel to and above the base of the dartboard and the π¦π¦-axis passing through themidpoint of the base. The π₯π₯-component of the base vertices are 3 and 3, and of the apex is 0. The centroidof a triangle occurs 2/3 of the way from a vertex to the mid
the 2020 MATHCOUNTS State Competition. These solutions provide creative and concise ways of solving the problems from the competition. There are certainly numerous other solutions that also lead to the correct answer, some even more creative and more concise! We encourage you to find a variety of approaches tosolving these fun and
math team placed fifth out of 32 middle schools competing in the MATHCOUNTS 2006 Oahu Chapter Competition, and are two of the six Oahu math teams that will represent Oahu in the Hawaii State MATHCOUNTS Competition on March 11, 2006. Other public . The first is the Sprint Round with its 30 problems, and then comes the Target
2005 MathCounts National Competition in Detroit, be it known that Detroit will host the 2005 Major League Baseball All-Star Game, the 2006 Super Bowl, and the 2009 NCAA Menβs Basketball Final Four. Obviously, the National MathCounts Competition fits right in with these other world-class c
The MATHCOUNTS Competition Program is designed to excite and challenge middle school students. With four levels of competition - school, chapter (local), state and national - the Competition Program provides students with the incentive to prepare throughout the school year to represent their schools at these MATHCOUNTS-hosted* events. MATHCOUNTS
2018 AMC 8 Problem 2 exactly the same as 2012 3 is MathCounts State Sprint Problem 3, and very similar to the following 4 problems: 2017 MathCounts Chapter Countdown #49 2016 MathCounts National Sprint #11 2016 - 2017 MathCounts School
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The MATHCOUNTS Competition Series promotes mathematics achievement through a series of engaging "bee" style contests. Like the Spelling Bee, MATHCOUNTS is a program where students progress from the School Competition, to the Local/Chapter Competition, to the State Competition, and ο¬nally to the National Competition.
2019 MATHCOUNTS State Competition. These solutions provide creative and concise ways of solving the problems from the competition. There are certainly numerous other solutions that also lead to the correct answer, some even more creative and more concise! We encourage you to find a variety of approaches to solving thesefun and challenging
This paper aims to extend this range and introduces a novel engineering application of Origami: Folded Textured Sheets. Existing applications of Origami in engineering can broadly be catego-rized into three areas. Firstly, many deployable structures take inspiration from, or are directly derived from, Origami folding. Examples are diverse and range from wrapping solar sails [Guest and .
